Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(2t^2+6,-t^2+t)$. What is the magnitude of the particle's velocity vector at $t=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $14$ (Choice B) B $10\sqrt{2}$ (Choice C) C $\sqrt{61}$ (Choice D) D $\sqrt{73}$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(2t^2+6,-t^2+t)$. We are asked to find the magnitude of the particle's velocity vector at $t=2$. In other words, we need to find $||\vec{v}(2)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(2t^2+6),\dfrac{d}{dt}(-t^2+t)\right) \\\\ &=(4t,-2t+1) \end{aligned}$ Finding $\vec{v}(2)$ $\begin{aligned} \vec{v}({2})&=(4({2}),-2({2})+1) \\\\ &=(8,-4+1) \\\\ &=(8,-3) \end{aligned}$ Finding $||\vec{v}(2)||$ $\begin{aligned} ||\vec{v}(2)||&=||(C{8},{-3})|| \\\\ &=\sqrt{(C{8})^2+({-3})^2} \\\\ &=\sqrt{73} \end{aligned}$ In conclusion, the magnitude of the particle's velocity vector at $t=2$ is $\sqrt{73}$.